Question

<?php
            include "conn.php";
            include "session.php";
        // Define $username and $password
            $n1=$_POST['Name'] ;
            $sql = "SELECT *FROM myDB.Mynew WHERE Fname like '%".$n1."%' ";
            $result = $conn->query($sql);
            if ( $result->num_rows >0) {
                    // output data of each row
                    echo '<table>';
                    echo '<tr>';
                    echo'<th>First Name</th>
                        <th>Last Name</th>
                        <th>Email</th>
                        <th>Password</th>';
                    echo'<tr>';
                    while($row = $result->fetch_assoc())  {
                        echo '<tr>';
                        echo '  <td>' . $row["firstname"] . '</td>';
                        echo '  <td>' . $row["lastname"] . '</td>';
                        echo '  <td>' . $row["email"] . '</td>';
                        echo '  <td>' . $row["password"] . '</td>';
                        echo '  </tr> ';                
                    }
                echo'</table>';
            }
            else {
                echo "<br> No Record Found to display";
            }
?>

当我运行此代码时，我收到以下通知：

注意：尝试在第14行获取非对象的属性 “if（$ result-＆gt; num_rows＆gt; 0）{”

请帮帮我。

Answer 1

更新表的结构时会发生此错误。即，更改表列名称或从当前用于检索数据的特定表中删除某些列。我想在你的表Mynew中删除任何列名字，姓氏，电子邮件，密码或者它们的名称是更新的

Answer 2

你需要在sql查询中添加一个空格......

尝试以下代码。它还会检查您的SQL查询是否返回结果：

            include "conn.php";
            include "session.php";
            // Define $username and $password
            $n1=$_POST['Name'] ;
            $sql = "SELECT * FROM myDB.Mynew WHERE Fname like '%".$n1."%' ";
            $result = $conn->query($sql);
            if($result){
            if ( $result->num_rows >0) {
                    // output data of each row
                    echo '<table>';
                    echo '<tr>';
                    echo'<th>First Name</th>
                        <th>Last Name</th>
                        <th>Email</th>
                        <th>Password</th>';
                    echo'<tr>';
                    while($row = $result->fetch_assoc())  {
                        echo '<tr>';
                        echo '  <td>' . $row["firstname"] . '</td>';
                        echo '  <td>' . $row["lastname"] . '</td>';
                        echo '  <td>' . $row["email"] . '</td>';
                        echo '  <td>' . $row["password"] . '</td>';
                        echo '  </tr> ';                
                    }
                echo'</table>';
            }
            else {
                echo "<br> No Record Found to display";
            }
            }else {
                echo "<br> Database error.";
            }

Answer 3

if ( isset($result->num_rows) && $result->num_rows >0) {

//--------

}

注意：尝试在if（$ result-＆gt; num_rows＆gt; 0）{获取非对象的属性

3 个答案: